Question

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.9 m/s at an angle of 20 ∘   above the horizontal. It is released 0.80 m above the floor.

What horizontal distance does the ball cover before bouncing?

Answer 1

Let us assume the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s2

Initial speed of the ball = V = 4.9 m/s

Angle the ball is thrown at = = 20 degrees

Initial horizontal velocity of the ball = Vx = VCos = (4.9)Cos(20) = 4.604 m/s

Initial vertical velocity of the ball = Vy = VSin = (4.9)Sin(20) = 1.676 m/s

Height from which the ball is released = H = 0.8 m

Time taken by the ball to reach the floor = T

When the ball bounces on the floor the vertical displacement of the ball is in the downwards direction and therefore negative.

T = 0.609 sec or -0.268 sec

Time cannot be negative.

T = 0.609 sec

Horizontal distance covered by the ball = R

There is no horizontal force acting on the ball therefore the horizonal velocity of the ball remains constant.

Horizontal distance the ball covers before bouncing = 2.8 m